Let's make the experiment more interesting...
--- Oleg Kobchenko <[EMAIL PROTECTED]> wrote:
> --- Raul Miller <[EMAIL PROTECTED]> wrote:
>
> > On 6/28/07, John Randall <[EMAIL PROTECTED]> wrote:
> > > X_i has the population distribution, so the variance of X_i is \sigma^2.
> > > The parenthetical (population variance) is an explanatory note, not an
> > > argument for \sigma^2.
> >
> > Ok, fair enough.
> >
> > That said, I believe I now understand your proof well enough to assert
> > that your proof works just as well if I use the same calculation
> > for sample standard deviation as I use for population standard
> > deviation. (With no "n-1" adjustment for the variance when working
> > with samples).
> >
> > In other words:
> >
> > Let $\mu$ and $\sigma$ be the population mean and standard deviation.
> > Fix the sample size $n$. Let $\bar X$ be the sample mean, $S^2$ the
> > sample variance given by
> >
> > $$ S^2=(1/n)\sum (X_i-\bar X)^2$$
> >
> > It is elementary to show $E(\bar X)=\mu$. We now show
> > $E(S^2)=\sigma^2$.
>
> This looks like Law of Large Numbers applied--
> however, not for baised estimate of the second
> central moment above. (See experiment.)
>
> ...
> >
> > I've checked my work on this -- I built myself a numerical
> > model representing most of the above statements and
> > have tested both versions of it [yours and mine] against
> > several different populations and sample sizes.
>
> It would be good to see the model.
> Here's a simple and straightforward Monte Carlo
> experiment running biased and unbiased estimates
> head-to-head.
... by throwing in another estimate of variance
with known population mean and 1/n.
> In fact it show that
>
> E((1/n)\sum (X_i-\bar X)^2) is (n-1)/n \sigma^2
> and
> E(1/(n-1)\sum (X_i-\bar X)^2) is \sigma^2
E((1/n)\sum (X_i-\mu)^2) is \sigma^2 -- known mean
NB. =========================================================
load 'stats'
samp=: ?@(# #) { ]
bvar=: var * <:@# % #
PS=: 100000
testvar=: 4 : 0
V=. B=. M=. 0
D=. normalrand PS
m=. mean D
for_i. i.x do.
d=. y samp D
V=. V + var d
B=. B + bvar d
M=. M + (+/*: d-m)%y
end.
(x%~V,B,M),bvar D
)
NB. =========================================================
10000 testvar 20 NB. V,M,D are almost same
0.995711 0.945926 0.995114 0.997888
0.945926 * 20%19 NB. compensate bias
0.995712
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